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## Interview MCQ Set 1

1. How a sharp separating point is occurs for solute and carrier in type 2 systems?
a) Extract reflux
b) Extract influx
c) Extract preflux
d) Detract influx

Answer: a [Reason:] With the help of extract relux, a sharp separating point is occurs for solute and carrier in type 2 systems.

2. For total reflux, difference points lie at:
a) Y=0
b) Y=-100 and Y=100
c) Y=0 and Y=infinity
d) Y=infinity and Y= -infinity

Answer: d [Reason:] Because F=B=D=0, for total reflux, the difference points lie at Y=infinity and Y= -infinity.

3. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.5 (YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 250

Answer: b [Reason:] Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 100*(0.5+0.5)-0.7/0.7-0.5=150.

4. Y which is ratio of mass of solvent to mass of solvent in liquid free phase: What does the value of Y depend on?
a) Viscosity of solvent
b) Amount of solvent
c) Mass fraction of solvent
d) Carrier solubility of solvent

Answer: d [Reason:] Y which is ratio of mass of solvent to mass of solvent in liquid free phase depends on carrier solubility of solvent.

5. Following the feed-stage location, the stripping stages are _______ until desired raffinate concentration is achieved.
a) Not done
b) Stepped on
c) Stepped off
d) Increased

Answer: c [Reason:] Following the feed-stage location, the stripping stages are stepped off until desired raffinate concentration is achieved.

6. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.5 (YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 200kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 300

Answer: d [Reason:] Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR, LR= 200*(0.5+0.5)-0.7/0.7-0.5=300.

7. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.4 (YD)
For extract leaving stage N is 0.6 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 40 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 250

Answer: a [Reason:] Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 100*(0.4+0.4)-0.6/0.6-0.4=100.

8. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.3(YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 125
c) 200
d) 250

Answer: b [Reason:] Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 250*(0.3+0.5)-0.7/0.7-0.5=125.

9. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.5 (YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 400
b) 500
c) 600
d) 700

Answer: c [Reason:] Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 400*(0.5+0.5)-0.7/0.7-0.5=600.

10. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.2 (YD)
For extract leaving stage N is 0.4 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 250

Answer: b [Reason:] Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR 100*(0.2+0.5)-0.4/0.4-0.2=150.

## Interview MCQ Set 2

1. Packed columns are better analyzed by:
a) Mass transfer coefficients
b) Equilibrium stage methods
c) Graphical methods
d) Algebraical methods

Answer: a [Reason:] Packed columns are better analyzed by mass transfer coefficients.

2. What is the formula for HETP or HETS?
a) HETP = l T *N T
b) HETP = l T-N T
c) HETP = l T /NT
d) HETP = lT+NT

Answer: c [Reason:] The formula for HETP is HETP = lT/NT where lT is the packed height length and the N T is the number of equilibrium stages.

3. At the interface of liquid and vapor, which interface exists?
a) Chemical
b) Physical
c) Thermal
d) No equilibrium exists

Answer: b [Reason:] At the interface of liquid and vapor, a physical equilibrium exists.

4. Which of the following is the correct illustration of rate of mass transfer?
a) r= ky a(y-y 1 ) =kx a(x-x 1 )
b) r= ky a(y 1 -y) =kx a(x 1 -x)
c) r= ky a(y-y 1 ) =kx a(x 1 -x)
d) r= a(y-y 1 ) =a(x 1 -x)

Answer: c [Reason:] r= k y a(y-y 1 ) =k xa(x 1 -x) is the correct illustration for rate of adsorption where kya and k x a are volumetric mass transfer coefficients.

5. Which of the following is the correct representation of overall mass transfer coefficient?
a) 1/Ky a = 1/ky a +K/Kx a
b) 1/Ky a = 1/ky a +K/Ky a
c) 1/Ky a = 1/ky a +1/KKx a
d) Ky a = 1/ky a +K/Kx a

Answer: a [Reason:] The correct representation of overall mass transfer coefficient is 1/K y a = 1/k y a +K/K x a, where K y is the overall mass transfer coefficient obtained from slope=point form.

6. What is the overall height of gas transfer unit if the volume is 1000m 3 and the overall gas phase coefficient is 2.6 and the inside cross sectional area is 20m 2 .
a) 12m
b) 13.6m
c) 18.9m
d) 19.2m

Answer: d [Reason:] HOG = V/ky aS, hence HOG= 1000/2.6*20=19.2.

7. What is the overall height of gas transfer unit if the volume is 780m 3 and the overall gas phase coefficient is 4.2 and the inside cross sectional area is 45m 2 .
a) 4.1m
b) 7.2m
c) 10.4m
d) 19.2m

Answer: a [Reason:] HOG = V/ky aS, hence HOG= 780/4.2*45=4.1.

8. What is the overall height of gas transfer unit if the volume is 489m 3 and the overall gas phase coefficient is 1.2 and the inside cross sectional area is 22m 2 .
a) 12m
b) 13.6m
c) 18.5m
d) 19.2m

Answer: c [Reason:] HOG = V/ky aS, hence HOG= 489/1.2*22=18.5.

9. What is the packed height length is HETP is 3ft and the number of equilibrium stages is
a) 0.75
b) 1.33
c) 12
d) 16

Answer: c [Reason:] HETP =lT /NT , hence lT =4*3=12.

10. What is the packed height length is HETP is 3.8ft and the number of equilibrium stages is
a) 0.76
b) 1.3157
c) 12
d) 19

Answer: c [Reason:] HETP =lT /NT , hence lT =3.8*5=19.

11. What is the packed height length is HETP is 6ft and the number of equilibrium stages is
a) 0.76
b) 1.3157
c) 30
d) 19

Answer: c [Reason:] HETP =lT /NT , hence lT =6*5=30.

12. What is the packed height length is HETP is 4.2ft and the number of equilibrium stages is
a) 0.76
b) 8.4
c) 12
d) 19

Answer: b [Reason:] HETP =lT /NT , hence lT =4.2*2=8.4.

## Interview MCQ Set 3

1. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.6h
c) 7.8h
d) 6.4h

Answer: a [Reason:] T= ρr2/6DeMbbCab, hence t=3.79h.

2. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.6M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.56h
c) 7.87h
d) 9.4h

Answer: b [Reason:] T= ρr2/6DeMbbCab, hence t=4.56h.

3. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 4.2h
b) 4.3h
c) 4.76h
d) 6.4h

Answer: c [Reason:] T= ρr2/6DeMbbCab, hence t=4.76h.

4. If the density of CuO in ore ρ = 0.08
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 6.08h
b) 6.32h
c) 6.35h
d) 6.4h

Answer: a [Reason:] T= ρr2/6DeMbbCab, hence t=6.08h.

5. If the density of CuO in ore ρ = 0.06
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 10.2h
b) 10.42h
c) 10.54h
d) 6.4h

Answer: b [Reason:] T= ρr2/6DeMbbCab, hence t=10.42h.

6. If the density of CuO in ore ρ = 0.03
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 2.1h
b) 2.28h
c) 7.8h
d) 6.4h

Answer: b [Reason:] T= ρr2/6DeMbbCab, hence t=2.28h.

7. If the density of CuO in ore ρ = 0.04
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.75h
c) 7.8h
d) 6.4h

Answer: b [Reason:] T= ρr2/6DeMbbCab, hence t=4.75h.

8. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 6.12h
b) 6.22h
c) 6.62h
d) 6.4h

Answer: c [Reason:] T= ρr2/6DeMbbCab, hence t=6.62h.

9. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 4.44*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 5.13h
b) 5.3h
c) 5.65h
d) 6.4h

Answer: a [Reason:] T= ρr2/6DeMbbCab, hence t=5.13h.

10. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 9*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 2.53h
b) 3.4h
c) 5.56h
d) 6.78h

Answer: a [Reason:] T= ρr2/6DeMbbCab, hence t=2.53h.

## Interview MCQ Set 4

1. Why is osmosis not a good separation process?
a) Because it involves semipermeable membrane
b) Because it transfers solvent from wrong direction
c) Because it involves movement of solute particles
d) Because it requires high temperatures

Answer: b [Reason:] Osmosis is not a good separation technique since it transfers solvent from wrong direction leading to mixing rather than separation.

2. What is the principle of reverse osmosis?
a) The direction of solvent flow Can be reversed by applying pressure greater than osmotic pressure
b) The direction of solvent flow can be reversed by applying pressure less than osmotic pressure
c) The direction of solvent flow can be reversed by using a permeable membrane
d) The solvent flow can be reversed by concentrating the mixture

Answer: a [Reason:] The direction of solvent flow Can be reversed by applying pressure greater than osmotic pressure. Hence P- osmotic pressure = pi.

3. If the Temperature is 300K and the concentration of solute(CB) is 0.5, what is the osmotic pressure?
a) 150R
b) 200R
c) 300R
d) 350R

Answer: a [Reason:] = RTCB, hence = r*300*0.5=150R.

4. If the Temperature is 400K and the concentration of solute(CB) is 0.5, what is the osmotic pressure?
a) 150R
b) 200R
c) 300R
d) 350R

Answer: b [Reason:] = RTCB, hence = r*400*0.5=200R.

5. If the Temperature is 100K and the concentration of solute(CB) is 0.6, what is the osmotic pressure?
a) 150R
b) 200R
c) 600R
d) 60R

Answer: d [Reason:] = RTCB, hence = R*100*0.6=60R.

6. If the Temperature is 350K and the concentration of solute(CB) is 0.6, what is the osmotic pressure?
a) 150R
b) 167R
c) 175R
d) 350R

Answer: c [Reason:] = RTCB, hence = R*350*0.5=175R.

7. If the Temperature is 300K and the concentration of solute(CB) is 0.75, what is the osmotic pressure?
a) 150R
b) 200R
c) 225R
d) 325R

Answer: c [Reason:] = RTCB, hence = R*300*0.75=225R.

8. If the Temperature is 300K and the concentration of solute(CB) is 0.3, what is the osmotic pressure?
a) 50R
b) 60R
c) 70R
d) 90R

Answer: d [Reason:] = RTCB, hence = R*300*0.3=90R.

9. Which of the following is not a use of reverse osmosis?
a) Treatment of industrial waste water to remove heavy metal ions
b) Separation of sulfites and bisulfites from paper and processing industry effluents
c) Treatment of municipal water to remove inorganic salts
d) Treatment of industrial effluents to decolorize it

Answer: d [Reason:] Reverse osmosis does not decolorize the water.

10. What is a normal tolerance value for concentration polarization coefficient?
a) 0.1
b) 0.2
c) 0.3
d) 0.5

Answer: b [Reason:] If he concentration polarization coefficient is greater than 0.2 then it is considered significant indicating changes in design.

## Interview MCQ Set 5

1. Which of the factors influence the feed conditions?
a) Phases
b) Capitol cost
c) Transport
d) Flow rate

Answer: d [Reason:] The ease of flow rate is a feed condition that will decide the separation process to be used.

2. Which of the factors influence the product conditions?
a) Phases
b) Capitol cost
c) Transport
d) Flow rate

Answer: a [Reason:] The phase of the products will also decide the product conditions which in turn decides the process used.

3. What are NOT the characteristics of separation operation?
a) Ease of staging
b) Ease of scale up
c) Operating cost
d) Energy requirements

Answer: c [Reason:] Operation cost is not a characteristic.

4. Which of the following can add significant cost to separation process?
a) Feed vaporization
b) Composition
c) Flow rate
d) Temperature

Answer: a [Reason:] The cost of feed vaporization is very high compared to the other three.

5. Which is the most important product condition?
a) Pressure
b) Temperature
c) Purity
d) Phases

Answer: c [Reason:] Purity is most important because other conditions can be altered by energy transfer.

6. What is the products price directly proportional to according to keller’s correlation?
a) Concentration
b) Dilution
c) Amount of feed
d) Amount of product

Answer: b [Reason:] According to Keller’s correlation, more the dilution more is the price of the protein.

7. Which of the following is true?
a) Operations based on barriers are more costly than operations based on use of solid agent
b) Operations based on use of solid agent are more costly than operations based on use of barriers
c) Operations based on use of creation are costlier than operation based on use of barriers
d) Operation based on addition of phase are more costlier than operations based on use of barriers

Answer: a [Reason:] Use of a barrier is most costly of all the other mentioned.

8. What is necessary if there is potential impurity or possible corrosion?
a) Parallel units
b) Maximum size units
c) Pilot-plant
d) New plant

Answer: c [Reason:] Operations near the bottom require pilot-plant test to see whether there are impurities or corrosions or not.

9. Ease of staging is not easy for which processes?
a) Distillation
c) Membrane separation
d) Crystallization